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13 July, 11:51

What is the ph of a solution that is 0.50 m in propanoic acid and 0.40 m in sodium propanoate. (ka for propanoic acid = 1.3 x 10-5) ?

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  1. 13 July, 12:06
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    The pH is given by the Henderson - Hasselbalch equation:

    pH = pKa + log ([A-]/[HA])

    pH = - log (1.3 x 10^-5) + log (0.50/0.40)

    pH = 4.98

    The answer to this question is 4.98.
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